Hashing

1. Hashing

2. The Search Problem • Find items with keys matching a given search key • Given an array A, containing n keys, and a search key x, find the index i such as x=A[i] • As in the case of sorting, a key could be part of a large record.

3. Applications • Keeping track of customer account information at a bank • Search through records to check balances and perform transactions • Keep track of reservations on flights • Search to find empty seats, cancel/modify reservations • Search engine • Looks for all documents containing a given word

4. Special Case: Dictionaries • Dictionary = data structure that supports mainly two basic operations: insert a new item and return an item with a given key • Queries: return information about the set S: • Search (S, k) • Minimum (S), Maximum (S) • Successor (S, x), Predecessor (S, x) • Modifying operations: change the set • Insert (S, k) • Delete (S, k) – not very often

5. Direct Addressing • Assumptions: • Key values are distinct • Each key is drawn from a universe U = {0, 1, . . . , m - 1} • Idea: • Store the items in an array, indexed by keys • Direct-address table representation: • An array T[0 . . . m - 1] • Each slot, or position, in T corresponds to a key in U • For an element x with key k, a pointer to x (or x itself) will be placed in location T[k] • If there are no elements with key k in the set, T[k] is empty, represented by NIL

6. Direct Addressing (cont’d)

7. Operations Alg.: DIRECT-ADDRESS-SEARCH(T, k) return T[k] Alg.: DIRECT-ADDRESS-INSERT(T, x) T[key[x]] ← x Alg.: DIRECT-ADDRESS-DELETE(T, x) T[key[x]] ← NIL • Running time for these operations: O(1)

8. Comparing Different Implementations • Implementing dictionaries using: • Direct addressing • Ordered/unordered arrays • Ordered/unordered linked lists Search Insert direct addressing O(1) O(1) ordered array O(N) O(lgN) ordered list O(N) O(N) unordered array O(1) O(N) unordered list O(1) O(N)

9. Examples Using Direct Addressing Example 1: Example 2:

10. Hash Tables • When K is much smaller than U, a hash table requires much less space than a direct-address table • Can reduce storage requirements to |K| • Can still get O(1) search time, but on the average case, not the worst case

11. Hash Tables Idea: • Use a function h to compute the slot for each key • Store the element in slot h(k) • A hash functionh transforms a key into an index in a hash table T[0…m-1]: h : U → {0, 1, . . . , m - 1} • We say that khashes to slot h(k) • Advantages: • Reduce the range of array indices handled: m instead of |U| • Storage is also reduced

12. Example: HASH TABLES 0 U (universe of keys) h(k1) h(k4) k1 K (actual keys) h(k2) = h(k5) k4 k2 k3 k5 h(k3) m - 1

13. Revisit Example 2

14. Do you see any problems with this approach? 0 U (universe of keys) h(k1) h(k4) k1 K (actual keys) h(k2) = h(k5) k4 k2 Collisions! k3 k5 h(k3) m - 1

15. Collisions • Two or more keys hash to the same slot!! • For a given set K of keys • If |K| ≤ m, collisions may or may not happen, depending on the hash function • If |K| > m, collisions will definitely happen (i.e., there must be at least two keys that have the same hash value) • Avoiding collisions completely is hard, even with a good hash function

16. Handling Collisions • We will review the following methods: • Chaining • Open addressing • Linear probing • Quadratic probing • Double hashing • We will discuss chaining first, and ways to build “good” functions.

17. Handling Collisions Using Chaining • Idea: • Put all elements that hash to the same slot into a linked list • Slot j contains a pointer to the head of the list of all elements that hash to j

18. Collision with Chaining - Discussion • Choosing the size of the table • Small enough not to waste space • Large enough such that lists remain short • Typically 1/5 or 1/10 of the total number of elements • How should we keep the lists: ordered or not? • Not ordered! • Insert is fast • Can easily remove the most recently inserted elements

19. Insertion in Hash Tables Alg.: CHAINED-HASH-INSERT(T, x) insert x at the head of list T[h(key[x])] • Worst-case running time is O(1) • Assumes that the element being inserted isn’t already in the list • It would take an additional search to check if it was already inserted

20. Deletion in Hash Tables Alg.: CHAINED-HASH-DELETE(T, x) delete x from the list T[h(key[x])] • Need to find the element to be deleted. • Worst-case running time: • Deletion depends on searching the corresponding list

21. Searching in Hash Tables Alg.: CHAINED-HASH-SEARCH(T, k) search for an element with key k in list T[h(k)] • Running time is proportional to the length of the list of elements in slot h(k)

22. T 0 chain m - 1 Analysis of Hashing with Chaining:Worst Case • How long does it take to search for an element with a given key? • Worst case: • All n keys hash to the same slot • Worst-case time to search is (n), plus time to compute the hash function

23. T n0 = 0 n2 n3 nj nk nm – 1 = 0 Analysis of Hashing with Chaining:Average Case • Average case • depends on how well the hash function distributes the n keys among the m slots • Simple uniform hashing assumption: • Any given element is equally likely to hash into any of the m slots (i.e., probability of collision Pr(h(x)=h(y)), is 1/m) • Length of a list: T[j] = nj, j = 0, 1, . . . , m – 1 • Number of keys in the table: n = n0 + n1 +· · · + nm-1 • Average value of nj: E[nj] = α = n/m

24. T 0 chain chain chain chain m - 1 Load Factor of a Hash Table • Load factor of a hash table T: = n/m • n = # of elements stored in the table • m = # of slots in the table = # of linked lists •  encodes the average number of elements stored in a chain •  can be <, =, > 1

25. Case 1: Unsuccessful Search(i.e., item not stored in the table) Theorem An unsuccessful search in a hash table takes expected time under the assumption of simple uniform hashing (i.e., probability of collision Pr(h(x)=h(y)), is 1/m) Proof • Searching unsuccessfully for any key k • need to search to the end of the list T[h(k)] • Expected length of the list: • E[nh(k)] = α = n/m • Expected number of elements examined in an unsuccessful search is α • Total time required is: • O(1) (for computing the hash function) + α 

26. Case 2: Successful Search

27. Analysis of Search in Hash Tables • If m (# of slots) is proportional to n (# of elements in the table): • n = O(m) • α = n/m = O(m)/m = O(1)  Searching takes constant time on average

28. Hash Functions • A hash function transforms a key into a table address • What makes a good hash function? (1) Easy to compute (2) Approximates a random function: for every input, every output is equally likely (simple uniform hashing) • In practice, it is very hard to satisfy the simple uniform hashing property • i.e., we don’t know in advance the probability distribution that keys are drawn from

29. Good Approaches for Hash Functions • Minimize the chance that closely related keys hash to the same slot • Strings such as pt and pts should hash to different slots • Derive a hash value that is independent from any patterns that may exist in the distribution of the keys

30. The Division Method • Idea: • Map a key k into one of the m slots by taking the remainder of k divided by m h(k) = k mod m • Advantage: • fast, requires only one operation • Disadvantage: • Certain values of m are bad, e.g., • power of 2 • non-prime numbers

31. Example - The Division Method m97 m 100 • If m = 2p, then h(k) is just the least significant p bits of k • p = 1  m = 2  h(k) = , least significant 1 bit of k • p = 2 m = 4  h(k) = , least significant 2 bits of k • Choose m to be a prime, not close to a power of 2 • Column 2: • Column 3: {0, 1} {0, 1, 2, 3} k mod 97 k mod 100

32. fractional part of kA = kA - kA The Multiplication Method Idea: • Multiply key k by a constant A, where 0 < A < 1 • Extract the fractional part of kA • Multiply the fractional part by m • Take the floor of the result h(k) = = m (k A mod 1) • Disadvantage: Slower than division method • Advantage: Value of m is not critical, e.g., typically 2p

33. Example – Multiplication Method

34. Universal Hashing • In practice, keys are not randomly distributed • Any fixed hash function might yield Θ(n) time • Goal: hash functions that produce random table indices irrespective of the keys • Idea: • Select a hash function at random, from a designed class of functions at the beginning of the execution

35. Universal Hashing (at the beginning of the execution)

36. Definition of Universal Hash Functions H={h(k): U(0,1,..,m-1)}

37. How is this property useful? Pr(h(x)=h(y))=

38. Universal Hashing – Main Result With universal hashing the chance of collision between distinct keys k and l is no more than the 1/mchance of collision if locations h(k) and h(l) were randomly and independently chosen from the set {0, 1, …, m – 1}

39. The class Hp,m of hash functions is universal Designing a Universal Class of Hash Functions • Choose a prime number p large enough so that every possible key k is in the range [0 ... p – 1] Zp = {0, 1, …, p - 1} and Zp* = {1, …, p - 1} • Define the following hash function ha,b(k) = ((ak + b) mod p) mod m,  a  Zp* and b Zp • The family of all such hash functions is Hp,m = {ha,b: a  Zp* and b  Zp} • a , b: chosen randomly at the beginning of execution

40. Example: Universal Hash Functions E.g.:p = 17, m = 6 ha,b(k) = ((ak + b) mod p) mod m h3,4(8) = ((38 + 4) mod 17) mod 6 = (28 mod 17) mod 6 = 11 mod 6 = 5

41. Advantages of Universal Hashing • Universal hashing provides good results on average, independently of the keys to be stored • Guarantees that no input will always elicit the worst-case behavior • Poor performance occurs only when the random choice returns an inefficient hash function – this has small probability

42. Open Addressing • If we have enough contiguous memory to store all the keys (m > N)  store the keys in the table itself • No need to use linked lists anymore • Basic idea: • Insertion: if a slot is full, try another one, until you find an empty one • Search: follow the same sequence of probes • Deletion: more difficult ... (we’ll see why) • Search time depends on the length of the probe sequence! e.g., insert 14

43. Common Open Addressing Methods • Linear probing • Quadratic probing • Double hashing • Note: None of these methods can generate more than m2different probing sequences!

44. Linear probing: Inserting a key • Idea: when there is a collision, check the next available position in the table (i.e., probing) h(k,i) = (h1(k) + i) mod m i=0,1,2,... • First slot probed: h1(k) • Second slot probed: h1(k) + 1 • Third slot probed: h1(k)+2, and so on • Can generate m probe sequences maximum, why? probe sequence: < h1(k), h1(k)+1 , h1(k)+2 , ....> wrap around

45. Linear probing: Searching for a key • Three cases: (1) Position in table is occupied with an element of equal key (2) Position in table is empty (3) Position in table occupied with a different element • Case 2: probe the next higher index until the element is found or an empty position is found • The process wraps around to the beginning of the table 0 h(k1) h(k4) h(k2) = h(k5) h(k3) m - 1

46. Linear probing: Deleting a key • Problems • Cannot mark the slot as empty • Impossible to retrieve keys inserted after that slot was occupied • Solution • Mark the slot with a sentinel value DELETED • The deleted slot can later be used for insertion • Searching will be able to find all the keys 0 m - 1

47. Primary Clustering Problem • Some slots become more likely than others • Long chunks of occupied slots are created  search time increases!! initially, all slots have probability 1/m Slot b: 2/m Slot d: 4/m Slot e: 5/m

48. Quadratic probing i=0,1,2,...

49. Double Hashing (1) Use one hash function to determine the first slot (2) Use a second hash function to determine the increment for the probe sequence h(k,i) = (h1(k) + i h2(k) ) mod m, i=0,1,... • Initial probe: h1(k) • Second probe is offset by h2(k) mod m, so on ... • Advantage: avoids clustering • Disadvantage: harder to delete an element • Can generate m2 probe sequences maximum

50. Double Hashing: Example h1(k) = k mod 13 h2(k) = 1+ (k mod 11) h(k,i) = (h1(k) + i h2(k) ) mod 13 • Insert key 14: h1(14,0) = 14 mod 13 = 1 h(14,1) = (h1(14) + h2(14)) mod 13 = (1 + 4) mod 13 = 5 h(14,2) = (h1(14) + 2 h2(14)) mod 13 = (1 + 8) mod 13 = 9 0 1 2 3 4 5 6 7 8 9 14 10 11 12